Question

PLEASE HELP!!! 15 PTS!!

Answer 1

Answer:

$\large\boxed{C.\ (2,\ -45^o)}$

Step-by-step explanation:

Regular coordinates (x, y)

Polar coordinates (r, φ)

$r=\sqrt{x^2+y^2}\\\\\psi=\arctan\frac{y}{x}$

We have the point

$(\sqrt2,\ -\sqrt2)$

Substitute:

$r=\sqrt{(\sqrt2)^2+(-\sqrt2)^2}=\sqrt{2+2}=\sqrt4=2\\\\\psi=\arctan\left(\frac{-\sqrt2}{\sqrt2}\right)=\arctan(-1)=-45^o$